CREATE TABLE: A table t is created with one column val of type INTEGER.

INSERT INTO: The values 1, 2, 2, 3, NULL, NULL, 4, 5 are inserted into the column val.

SELECT COUNT(val): This query counts all non-NULL values in the val column.

Total non-NULL values: 1, 2, 2, 3, 4, 5

Result: 6

SELECT COUNT(DISTINCT val): This query counts only the unique non-NULL values in the val column.

Unique values: 1, 2, 3, 4, 5

Result: 5

#include<stdio.h>
using namespace std;

// Recursive function to check if a string is a palindrome
bool isPalindrome(string str, int start, int end) {
    // Base case: If the start index is greater than or equal to the end index
    if (start >= end) {
        return true;
    }
    // Check if the first and last characters are the same
    if (str[start] != str[end]) {
        return false;
    }
    // Recursive case: Check the substring excluding the first and last characters
    return isPalindrome(str, start + 1, end - 1);
}

int main() {
    string input;
    cout << "Enter a string: "; cin >> input;

    if (isPalindrome(input, 0, input.length() - 1)) {
        cout << input << " is a palindrome." << endl;
    } else {
        cout << input << " is not a palindrome." << endl;
    }

    return 0;
}

Sample I/O:
Enter a string: noon
noon is a palindrome.
=== Code Execution Successful ===

🔗 Run Online: Check Palindrome

Capacity of a Track

Bytes per Track = Bytes per Sector × Sectors per Track
Bytes per Track = 512 × 50 = 25,600 bytes

Capacity of a Surface

Bytes per Surface = Bytes per Track × Tracks per Surface
Bytes per Surface = 25,600 × 2000 = 51,200,000 bytes (50,000 KB)

Capacity of the Disk

Bytes per Disk = Bytes per Surface × Number of Surfaces
Number of Surfaces = 5 Platters × 2 Surfaces per Platter = 10 Surfaces
Bytes per Disk = 51,200,000 × 10 = 512,000,000 bytes (500,000 KB)

C = B log₂(1 + SNR)
= 3000 log₂(1 + 3162)
= 3000 log₂(3163)
= 3000 × 11.62
= 34,860 bps

Web Caching

Web caching is a technique where frequently accessed web content (such as web pages, images, or files) is stored temporarily in a cache to reduce server load and improve response time.

Is web caching available for every user request?

No, web caching is not available for every user request.

Reasons:

• First-time Requests: If a user requests a resource for the first time, it is not available in the cache and must be fetched from the original server.
• Dynamic Content: Pages that change frequently (such as personalized dashboards or live data) are usually not cached.
• Cache Expiration: Cached content may expire after a certain time (TTL), requiring a fresh copy from the server.
• Cache-Control Rules: Some servers specify that certain content should not be cached for security or accuracy reasons.

Conclusion: Web caching improves performance, but it is applied only when cached content is valid and available.

ARQ (Automatic Repeat reQuest)

ARQ is an error control technique used in data communication to ensure reliable data transmission. It detects errors in transmitted frames and automatically requests retransmission of incorrect or lost data.

Working Principle:

• The sender transmits a data frame to the receiver.
• The receiver checks the frame for errors using error-detection techniques.
• If the frame is correct, the receiver sends an ACK (Acknowledgement).
• If the frame is incorrect or lost, the receiver sends a NACK or no response.
• The sender retransmits the frame when required.

Types of ARQ:

• Stop-and-Wait ARQ
• Go-Back-N ARQ
• Selective Repeat ARQ

         a
       /   \
      b     d
     / \   / \
    e   f g   i
   / \       / \
  j   k     l   m
     / \
    n   o
         \
          p

Subnet Mask: /22 corresponds to the subnet mask 255.255.252.0.

Block Size: 2^(32−22)=2^10=1024.

Network Address: 172.16.16.0/22.

Broadcast Address: 172.16.19.255/22.

Total Valid Hosts: 2^(32−22)−2=1024−2=1022

IP:     10101100.00010000.00010000.10001001
Mask:   11111111.11111111.11111100.00000000
------------------------------------------------
Result: 10101100.00010000.00010000.00000000

Network (Binary):
10101100.00010000.00010000.00000000

Host bits (10 bits) = 1111111111

Broadcast (Binary):
10101100.00010000.00010011.11111111

1. Session Hijacking

Problem: Session hijacking is an attack where an attacker steals a valid user session ID to gain unauthorized access to a web application. This can occur through packet sniffing, XSS attacks, or insecure cookies.

Possible Solutions:

• Use HTTPS to encrypt data transmission.
• Implement secure and HttpOnly cookies.
• Regenerate session IDs after login.
• Set proper session timeout.

2. SQL Injection

Problem: SQL injection occurs when an attacker inserts malicious SQL queries through input fields, allowing unauthorized access to the database or manipulation of data.

Possible Solutions:

• Use prepared statements and parameterized queries.
• Validate and sanitize user inputs.
• Apply least privilege principle for database users.
• Use Web Application Firewalls (WAF).

How HTTP Works

HTTP (HyperText Transfer Protocol) is a request–response protocol used for communication between a web client (browser) and a web server.

• The client sends an HTTP request (e.g., GET, POST) to the server.
• The server processes the request.
• The server sends back an HTTP response containing status code and data.
• The browser renders the received content.

How many HTTP requests are needed?

The number of HTTP requests depends on the number of resources a webpage contains.

• 1 request for the main HTML page.
• Each embedded resource (image, CSS, JS, video) needs a separate request.

Example:

If a webpage contains:

• 1 HTML file
• 2 images
• 1 CSS file
• 1 JavaScript file

Total HTTP requests = 5

Given Function:

Slow_solution(n){
 result = 0;
 for i = 1 to n{
   for j = 1 to (i + 1){
     result += 1;
   }
 }
 return result;
}

Step-by-step Time Complexity Analysis:

• The outer loop runs from i = 1 to n, so it executes n times.
• The inner loop runs from j = 1 to (i + 1).
• For each value of i, the inner loop runs approximately i times.

Total number of operations:

1 + 2 + 3 + ... + n = n(n + 1)/2

This is a quadratic expression.

Final Time Complexity:

O(n²)